$#Muzan$

`a.)`

`A=((2sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)/(3-sqrtx)).(x-3sqrtx)/(xsqrtx+1)`

`ĐK:xne4 ,xne9,x>=0`

`A=((2sqrtx-9)/((sqrtx-2).(sqrtx-3))-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)/(sqrtx-3)).(sqrtx .(sqrtx-3))/(xsqrtx+1)`

`A=(2sqrtx-9-(sqrtx+3)(sqrtx-3)+(2sqrtx+1)(sqrtx-2))/((sqrtx-2)(sqrtx-3)). (sqrtx(sqrtx-3))/(xsqrtx+1)`

`A=(2sqrtx-9-x+9+2x-3sqrtx-2)/(sqrtx-2) . (sqrtx)/(xsqrtx+1)`

`A=(x-sqrtx-2)/(sqrtx-2) . (sqrtx)/((sqrtx+1)(x-sqrtx+1))`

`A=((sqrtx-2)(sqrtx+1))/(sqrtx-2) . (sqrtx)/((sqrtx+1)(x-sqrtx+1))`

`A=(sqrtx)/(x-sqrtx+1)`

Vậy `A=(sqrtx)/(x-sqrtx+1)` tại `xne4,xne9,x>=0`

`b.)`

Ta có `x=sqrt(3+2sqrt2)+sqrt(11-6sqrt2)+12`

`=sqrt((sqrt2+1)^2) +sqrt((3-sqrt2)^2)+12`

`=|sqrt2+1|+|3-sqrt2|+12`

`=sqrt2+1+3-sqrt2+12`

`=16`

Tại `x=16` TM ĐKXĐ

`A=(sqrt16)/(16-sqrt16+1)=4/13`

Vậy tại `x=16` thì `A=4/13`

`c.)`

Ta có `A-1=(sqrtx)/(x-sqrtx+1)-1`

`=(sqrtx-x+sqrtx-1)/(x-sqrtx+1)`

`=-(x-2sqrtx+1)/(x-sqrtx+1)`

`=-((sqrtx-1)^2)/(x-sqrtx+1)`

Mà `-(sqrtx-1)^2<=0AAx`

`x-sqrtx+1=(sqrtx-1/2)^2+3/4>=3/4>0AAx`

Hay `-((sqrtx-1)^2)/(x-sqrtx+1)<=0`

Hay `A-1<=0`

Hay `A<=1`

`Max_A=1` khi `sqrtx-1=0=>sqrtx=1=>x=1` ( Thỏa mãn )

Vậy `Max_A=1` khi `x=1`